The magnetic flux near the axis and inside the air-core solenoid of length 60 cm carrying current I is 1.57× 106 Wb. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ 0=4π × 10- 7 SI units)

Question

The magnetic flux near the axis and inside the air-core solenoid of length 60 cm carrying current I is 1.57× 106 Wb. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ 0=4π × 10- 7 SI units) (A) 0.25 A m2 (B) 0.5 A m2 (C) 0.75 A m2 (D) 1 A m2

Tardigrade Admin · Accepted Answer

Magnetic induction inside the solenoid B=(μ 0 N I/L) Magnetic flux, φ=BA =(μ 0 N I A/L) Magnetic moment =NIA=(φ L/μ 0) =(1.57 × 10- 6 × 0.6/4 × 3.14 × 10- 7) =0.75 Am2

Q. The magnetic flux near the axis and inside the air-core solenoid of length $60 c m$ carrying current $I$ is $1.57 \times 1 0^{6} Wb .$ Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, $μ_{0} = 4 π \times 1 0^{- 7}$ SI units)

$0.25 A m^{2}$

$0.5 A m^{2}$

$0.75 A m^{2}$

$1 A m^{2}$

Magnetic induction inside the solenoid
$B = \frac{μ _{0} N I}{L}$
Magnetic flux, $ϕ = B A$
$= \frac{μ _{0} N I A}{L}$
Magnetic moment $= N I A = \frac{ϕ L}{μ _{0}}$
$= \frac{1.57 \times 1 0 ^{- 6} \times 0.6}{4 \times 3.14 \times 1 0 ^{- 7}}$
$= 0.75 A m^{2}$

Q. The magnetic flux near the axis and inside the air-core solenoid of length 60cm carrying current I is 1.57×106Wb. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ0​=4π×10−7 SI units)

0.25Am2

0.5Am2

0.75Am2

1Am2